Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $z \neq 0$. $y = \dfrac{z^2 + 12z + 35}{z + 5} \times \dfrac{-9z - 90}{9z + 63} $
Solution: First factor the quadratic. $y = \dfrac{(z + 7)(z + 5)}{z + 5} \times \dfrac{-9z - 90}{9z + 63} $ Then factor out any other terms. $y = \dfrac{(z + 7)(z + 5)}{z + 5} \times \dfrac{-9(z + 10)}{9(z + 7)} $ Then multiply the two numerators and multiply the two denominators. $y = \dfrac{ (z + 7)(z + 5) \times -9(z + 10) } { (z + 5) \times 9(z + 7) } $ $y = \dfrac{ -9(z + 7)(z + 5)(z + 10)}{ 9(z + 5)(z + 7)} $ Notice that $(z + 5)$ and $(z + 7)$ appear in both the numerator and denominator so we can cancel them. $y = \dfrac{ -9\cancel{(z + 7)}(z + 5)(z + 10)}{ 9(z + 5)\cancel{(z + 7)}} $ We are dividing by $z + 7$ , so $z + 7 \neq 0$ Therefore, $z \neq -7$ $y = \dfrac{ -9\cancel{(z + 7)}\cancel{(z + 5)}(z + 10)}{ 9\cancel{(z + 5)}\cancel{(z + 7)}} $ We are dividing by $z + 5$ , so $z + 5 \neq 0$ Therefore, $z \neq -5$ $y = \dfrac{-9(z + 10)}{9} $ $y = -(z + 10) ; \space z \neq -7 ; \space z \neq -5 $